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Tough Kinematics Question

  1. Sep 12, 2008 #1
    ok i need help with this type of question

    What are the (a)x and (b)y components of the vector that must be added to the following three vectors, so that the sum of the four vectors is zero? Due east is the +x direction, and due north is the +y direction.

    A= 141 units, 59.0 ° south of west
    B= 243 units, 38.0 ° south of east
    C= 173 units, 29.0 ° north of east

    a)=___________
    b)=___________

    i tried finding the x components by simple multiplying the units by cos(angle), found my answer to be 270 units which i know is wrong.

    I have a feelin that yur supposed go x=sin n y=cos in some vectors but im not 100% sure
     
  2. jcsd
  3. Sep 12, 2008 #2
    Very Tough Kinematics 1 dimension problem

    1. The problem statement, all variables and given/known data

    What are the (a)x and (b)y components of the vector that must be added to the following three vectors, so that the sum of the four vectors is zero? Due east is the +x direction, and due north is the +y direction.

    A= 141 units, 59.0 ° south of west
    B= 243 units, 38.0 ° south of east
    C= 173 units, 29.0 ° north of east

    a)=___________
    b)=___________

    Answer must be in 3 significant digits

    2. Relevant equations



    3. The attempt at a solution

    The answer is not a):270 or b) -204.95 ive tried n its wrong...

    so please if u can help out than great
     
  4. Sep 12, 2008 #3

    danago

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    Re: Very Tough Kinematics 1 dimension problem

    Well the first thing i would do is decompose each of the given vectors into their x and y components. Is that how you did it? How exactly did you get the answers you came up with?
     
    Last edited: Sep 12, 2008
  5. Sep 12, 2008 #4
    Re: Very Tough Kinematics 1 dimension problem

    thats exactly how i started

    A= x: 173cos29=151.30
    y: 173sin29= 83.87

    B= x: 243cos38=191.48
    y: 243sin38= 149.60

    C= x: 141cos59= 72.62
    y: 141sin59= 138.35

    when i looked for the displacement of x i got 270.16, and to signifiant digits i cant get 3 out of that cuz the answer equires 3
    for y the total displacement : was -204.08 and that was wrong as well....

    I dont know i think this is one of those types of questions where some vectors you have to switch between x n y, like x wud be sin and y wud be cos, but i just dont know when and how to do that in the question...
     
  6. Sep 12, 2008 #5

    gabbagabbahey

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    Re: Very Tough Kinematics 1 dimension problem

    If you add 270 to 270 do you really get zero?:wink:
     
  7. Sep 12, 2008 #6

    danago

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    Re: Very Tough Kinematics 1 dimension problem

    Something you seem to have overlooked is the positive and negative signs on some of your components. It often makes it easier to draw each vector in the x-y plane first.

    Looking at vector B, for example. 243 units, 38.0 ° south of east. If you draw that, it will look like a vector with a length of 243 units, and it will point in a direction of 38 degrees below the positive x axis. Now think about each of its components; Its x component points in the +x direction so it will be positive, but the y component is in the -y direction and so it will be negative,
     
  8. Sep 12, 2008 #7
    Re: Very Tough Kinematics 1 dimension problem

    ^^^ i actually have done, although i didnt display it,

    for total x value : A + B - C , C is the only one in the west direction therfore I had to subtract it

    (151.30) + (191.48)- (72.62) = 270. 15

    i did the same for y but its the wrong answer...
     
  9. Sep 12, 2008 #8

    gabbagabbahey

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    Re: Very Tough Kinematics 1 dimension problem

    The original question ask you which vector you have to add to this total in order to get zero....if you add 270 to 270 do you get zero?:wink:
     
  10. Sep 12, 2008 #9
    Choose a direction to call 0°, and express the angles with respect to that reference.

    [edit: look at the problem statement for clues as to where 0 and 90 should point]

    Regards,

    Bill
     
    Last edited: Sep 12, 2008
  11. Sep 12, 2008 #10
    ? come again? lol sorry i misunderstand
     
  12. Sep 12, 2008 #11
    Re: Very Tough Kinematics 1 dimension problem

    ^^ yea if the answer was 270 and i add (-270) that will become 0, only problem is i have to display the answer in 3 significant digits, and -270 is only 2 significant digits....
     
  13. Sep 12, 2008 #12

    gabbagabbahey

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    Re: Very Tough Kinematics 1 dimension problem

    -270 is 3 sig digs
     
  14. Sep 12, 2008 #13
    Re: Very Tough Kinematics 1 dimension problem

    i believe its 2, becuz if u were to put it into scientific notation it wud 27 x 10 ^ 1 or 2.7 X 10 ^ 2. A Trailing 0 is only significant if its after a decimal point, 27.0 wud be 3, but thats not the answer
     
  15. Sep 12, 2008 #14

    gabbagabbahey

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    Re: Very Tough Kinematics 1 dimension problem

    No, in this case the zero is significant because it was rounded from 0.1 to 0 and so you would have -2.70 X 10^2...Just because the 3rd sig fig is zero doesn't mean it isn't significant.
     
  16. Sep 12, 2008 #15
    59° South of West is either 239° from East, or -121° from East.

    Regards,

    Bill
     
  17. Sep 13, 2008 #16
    I cannot figure this out at all, its a question on the computer so when i type in -270, it tells me, please put the correct number of significant figures and i cant display it -2.70 x 10 ^ 2, cuz it doesnt allow scientic notation...
     
  18. Sep 13, 2008 #17

    gabbagabbahey

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    That's strange, -270 should be the right answer...maybe try -269 or -271 in case the instructor accidentally rounded each component off before adding them together.

    Are you sure you copied the question out exactly?
     
  19. Sep 13, 2008 #18
    im positive^^^
     
  20. Sep 13, 2008 #19

    gabbagabbahey

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    Well then I recommend you talk to /email your instructor, show him your solution and ask if it is correct
     
  21. Sep 13, 2008 #20
    lol yur a genius im an idiot^^, the answer was -269 for x and for y it was +186.59, i tried typing it in and it was correct thanks alot man:)
     
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